Is a weakly$^*$ convergent net of positive(!) Radon measures eventually norm-bounded?

Question

Let $X$ be a locally compact Hausdorff space and $C_0(X)$ the space of continuous functions vanishing at infinity.

If $\mu_n$ is a sequence of positive Radon measures on $X$ such that $\mu_n \to \mu$ for the weak$^*$ topology $\sigma(M(X), C_0(X))$ then $\mu_n$ is norm-bounded.

How does this generalize to nets? In contrast to sequences, a convergent net need not be bounded ("the initial parts of a net can be infinitely long"). Moreover, a weakly$^*$ convergent net (of signed Radon measures) need not be eventually bounded, see here. Does positivity help out? If $\mu_\alpha \geq 0$ and $\mu_\alpha \to \mu$ for the weak$^*$ topology, does it follow that $\mu_\alpha$ is eventually norm-bounded, i.e. is there $\alpha_0$ such that $\sup_{\alpha \geq \alpha_0} \lVert \mu_\alpha \rVert = \sup_{\alpha \geq \alpha_0} \mu_\alpha(X) < \infty$?

This is related to the question here, but may be of independent interest for the MSE community.

Edit: Note that this is true if $X$ is compact, because $1_X \in C_0(X) = C(X)$. Then $\mu_\alpha \to \mu$ weakly$^*$ implies $\mu_\alpha(X) \to \mu(X)$, so that $\mu_\alpha(X)$ is a convergent net in $\mathbb{R}$ and therefore eventually bounded: for $\varepsilon = 1$ there is $\alpha_0$ such that $0 \leq \mu_\alpha(X) \leq \mu(X) + 1$ for all $\alpha \geq \alpha_0$. In particular, this example of an unbounded weakly$^*$ convergent net of Radon measures does not work for positive measures/functionals.

Answer 1

For $X = [0, \omega_1)$ ($\omega_1$ the first uncountable ordinal), a $\sigma(M(X), C_0(X))$-convergent net of positive Radon measures need not be eventually uniformly bounded. The construction is similar to this one.

Let me first provide some general remarks. For $\mu \in M(X)$ and $f \in C_0(X)$ denote by $\mu f := \int f \, d\mu$. A neighborhood base of $0$ in $M(X)$ for the considered weak$^*$ topology $\sigma(M(X), C_0(X))$ is given by the sets $V_{f_1, \dots, f_n} = \{ \mu \in M(X) \mid |\mu f_1| < 1, \dots, |\mu f_n| < 1 \}$ where $f_1, \dots, f_n \in C_0(X)$. Therefore, $U_{f_1, \dots, f_n} := V_{f_1, \dots, f_n} \cap M(X)^+$ is a neighborhood base of $0$ on the set of positive measures $M(X)^+$. For $f_1, \dots, f_n \in C_0(X)$ set $f := |f_1| + \dots + |f_n| \in C_0(X)^+$. Then $U_f = \{ \mu \in M(X)^+ \mid \mu f < 1 \} \subseteq U_{f_1, \dots, f_n}$ since $|\mu f_i| \leq \mu |f_i| \leq \mu f$. Therefore, the sets $U_f$, $f \in C_0(X)^+$ form a neighborhood base of $0$ in $M(X)^+$.

On $X = [0, \omega_1)$ every continuous function is eventually constant. Therefore, any $f \in C_0(X)$ has compact support. For any $\alpha < \omega_1$ the set $[0, \alpha] \subseteq X$ is compact and any compact set of $X$ is contained in some $[0, \alpha]$. Consider the index set $\mathcal{F} := C_0(X)^+ \times \mathbb{N}$ directed by $(f, n) \preceq (g, m) :\Leftrightarrow f \leq g$ pointwise. Fix $f \in \mathcal{F}$. Since $f$ has compact support, there is $x \in [0, \omega_1)$ such that $f(x) = 0$. Therefore, $\delta_x \in U_f$ and moreover $n \delta_x \in U_f$ for any $n \in \mathbb{N}$. Set $\mu_{(f,n)} := n \delta_x \in M(X)^+$. Then the net $(\mu_{(f,n)})_{(f, n) \in \mathcal{F}}$ converges to $0$. In fact, if $U$ is an open neighborhood of $0$ in $M(X)^+$ then there is $f_0 \in C_0(X)^+$ such that $U_{f_0} \subseteq U$. Then for all $(f, n) \succeq (f_0, 0)$ (i.e. for all $f \geq f_0$) it holds $\mu_{(f,n)} \in U_f \subseteq U_{f_0} \subseteq U$. But the net $\mu_{(f,n)}$ is not eventually uniformly bounded: for any $(f_0, n_0)$ and any $r \geq 0$ there is $n \in \mathbb{N}$, $n \geq r$ such that $\lVert \mu_{(f_0,n)} \rVert = n \geq r$, hence $\sup_{(f,n) \succeq (f_0,n_0)} \lVert \mu_{(f,n)} \rVert = \infty$.